Problem: Let $y=(1+5x-x^2)^{^{{\scriptsize\dfrac14}}}$. Evaluate $\dfrac{dy}{dx}$ at $x=5$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac54$ (Choice B) B $1$ (Choice C) C $-5$ (Choice D) D $\dfrac14$
Explanation: Let's start by finding the expression for $\dfrac{dy}{dx}$. Then, we can evaluate it at $x=5$. $(1+5x-x^2)^{^{{\scriptsize\dfrac14}}}$ is a power function with a rational exponent, but its argument isn't simply $x$. Therefore, it defines a composite power function. In other words, suppose $u(x)=1+5x-x^2$, then $y=[u(x)]^{^{{\scriptsize\dfrac14}}}$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[[u(x)]^{^{{\scriptsize\dfrac14}}}\right]=\dfrac{1}{4}[u(x)]^{^{-\scriptsize\dfrac{3}{4}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(1+5x-x^2)^{^{{\scriptsize\dfrac14}}} \\\\ &=\dfrac{d}{dx}[u(x)]^{^{{\scriptsize\dfrac14}}}&&\gray{\text{Let }u(x)=1+5x-x^2} \\\\ &=\dfrac{1}{4}[u(x)]^{^{-\scriptsize\dfrac{3}{4}}}u'(x) \\\\ &=\dfrac{1}{4}[1+5x-x^2]^{^{-\scriptsize\dfrac{3}{4}}}(5-2x)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $x= 5$. $\begin{aligned} &\phantom{=}\dfrac{1}{4}\Bigl(1+5( 5)-( 5)^2\Bigr)^{^{-\scriptsize\dfrac{3}{4}}}\cdot\Bigl(5-2( 5)\Bigr) \\\\ &=\dfrac{1}{4}\cdot 1^{^{-\scriptsize\dfrac{3}{4}}}\cdot(-5) \\\\ &=\dfrac{1}{4}\cdot 1\cdot (-5) \\\\ &=-\dfrac{5}{4} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=5$ is $-\dfrac{5}{4}$.